JEE MAIN - Physics (2015 (Offline) - No. 2)
A red $$LED$$ emits light at $$0.1$$ watt uniformly around it. The amplitude of the electric field of the light at a distance of $$1$$ $$m$$ from the diode is :
$$5.48$$ $$V/m$$
$$7.75$$ $$V/m$$
$$1.73$$ $$V/m$$
$$2.45$$ $$V/m$$
Explanation
Intensity of light at a distance r, $$I = {P \over {4\pi {r^2}}}$$ [P = power]
Again, if the amplitude of the electric field is E0
then $$I = {1 \over 2}c{ \in _0}E_0^2$$
$$\therefore$$ $${P \over {4\pi {r^2}}} = {1 \over 2}c{ \in _0}E_0^2$$
or, $${E_0} = \sqrt {{P \over {2\pi c{ \in _0}{r^2}}}} $$
$$ = \sqrt {{{0.1} \over {2 \times 3.14 \times (3 \times {{10}^8}) \times (8.85 \times {{10}^{ - 12}}) \times {1^2}}}} $$
= 2.45 V/m
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